Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Gọi \(A=\frac{1005}{2011}\)
A=1/3 + 1/3.5 + 1/5.7 +...............+1/x.(x+2)
A=1/1.3 + 1/3.5 + 1/5.7 +...............+1/x.(x+2)
A . 2=2/1.3 + 2/3.5 + 2/5.7 +......................+2/x.(x+2)
A . 2=1/1-1/3+1/3-1/5+1/5-1/7+..............+1/x-1/x+2
A . 2=1/1+(1/3-1/3)+(1/5-1/5)+..............+(1/x-1/x)-1/x+2
A . 2=1/1-1/x+2
Suy gia:1005/2011 . 2=1/1-1/x+2
2010/2011 =1/1-1/x+2
1/x+2 =1/1-2010/2011
1/x+2 =1/2011
Suy gia:x+2=2011
x =2011-2
x =2009
\(S=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\)
\(2S=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
\(2S=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(2S=\dfrac{1}{3}-\dfrac{1}{99}\)
\(2S=\dfrac{32}{99}\)
\(S=\dfrac{32}{99}:2\)
\(S=\dfrac{16}{99}\)
Cảm ơn bạn rất nhiều! Bạn đã cứu mình rùi! Xin trân thành cảm ơn!
Ta có : \(S=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(\Rightarrow2S=2\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\right)\)
\(\Rightarrow2S=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(\Rightarrow2S=\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow2S=\left(\frac{1}{3}-\frac{1}{99}\right)+\left[\left(\frac{1}{5}+...+\frac{1}{97}\right)-\left(\frac{1}{5}+\frac{1}{7}+...+\frac{1}{97}\right)\right]\)
\(\Rightarrow2S=\left(\frac{33}{99}-\frac{1}{99}\right)+0\)
\(\Rightarrow2S=\frac{32}{99}\)
\(\Rightarrow S=\frac{32}{99}\div2\)
\(\Rightarrow S=\frac{16}{99}\)
\(S=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{97.99}\)
\(S=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(S=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(S=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(S=\frac{1}{2}.\frac{32}{99}=\frac{16}{99}\)
\(\dfrac{1}{1.3}\)+ \(\dfrac{1}{3.5}\)+ \(\dfrac{1}{5.7}\)+....+\(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{1005}{2011}\)
1- \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\)+ \(\dfrac{1}{5}\)- \(\dfrac{1}{7}\)+....+\(\dfrac{1}{x}\)- \(\dfrac{1}{x+1}\)= \(\dfrac{1005}{2011}\)
1- \(\dfrac{1}{x+1}\)= \(\dfrac{1005}{2011}\)
\(\dfrac{1}{x+1}\)= 1- \(\dfrac{1005}{2011}\)
\(\dfrac{1}{x+1}\)= \(\dfrac{1006}{2011}\)
=> x +1= 2011
=> x= 2011-1
=> x=2010
Bài này mk lm đại nha bn 
! Cs j sai mong bn bỏ qua 
.
\(1-\frac{1}{x+20}=\frac{1005}{2011}\)
\(\frac{1}{x+20}=1-\frac{1005}{2011}\)
\(\frac{1}{x+20}=\frac{1006}{2011}\)
=>1006.(x+20)=2011.1
=>1006x+20120=2011
1006x=2011-20120=-18109
x=\(\frac{-18109}{1006}\)
\(1-\frac{1}{\left(x+20\right)}=\frac{1005}{2011}\)
\(\Rightarrow\frac{1}{\left(x+20\right)}=1-\frac{1005}{2011}=\frac{1006}{2011}\)
\(\Rightarrow1:\left(x+20\right)=\frac{1006}{2011}\)
\(\Rightarrow x+20=\frac{2011}{1006}\)
\(\Rightarrow x=\frac{2011}{1006}-20=\frac{-18109}{1006}\)
a, 2/3 của -420 là :
-420 x 2/3 = -280
Số cần tìm là :
-280 x 5/8 = -175
Vậy số cần tìm là -175
b, 1/1.3 + 1/3.5 + 1/5.7 + ... + 1/x ( x + 2 ) = 1005 / 2011
1/2 x ( 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/ ( x ( x + 2 ) = 1005 / 2011
1/2 x ( 1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/x - 1/ x + 2 ) = 1005 / 2011
1/2 x ( 1 - 1/ x + 2 ) = 1005 / 2011
1 - 1 / x + 2 = 1005 / 2011 : 1/2
1 - 1 / x + 2 = 2010 / 2011
x + 2 / x + 2 - 1 / x + 2 = 2010 / 2011
x + 2 - 1 / x + 2 = 2010 / 2011
x + 1 / x + 2 = 2010 / 2011
+> x + 1 = 2010
x = 2010 - 1
x = 2009
+> x + 2 = 2011
x = 2011 - 2
x = 2009
Vậy x = 2009
Tk nha Đúng đó !!
=>2/1*3+2/3*5+...+2/(2x-1)(2x+1)=98/99
=>1-1/3+1/3-1/5+...+1/(2x-1)-1/(2x+1)=98/99
=>1-1/(2x+1)=98/99
=>1/(2x+1)=1/99
=>2x+1=99
=>x=49
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{1005}{2011}\)
\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}\right)=\frac{1005}{2011}\)
\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\frac{1}{3}-\frac{1}{x+2}=\frac{2010}{2011}\)
\(\frac{1}{x+2}=\frac{1}{3}-\frac{2010}{2011}\)
\(\frac{1}{x+2}=\frac{1}{2011}\)
\(\Rightarrow x+2=2011\)
\(x=2009\)
Đặt biểu thứ là A
2A=2/1.3+2/2.5+...+2/x.x+2
2A=1-1/3+1/3-1/5+.......+1/x-1/x+2
2A=1-1/x+2